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$$ xu_{xy}+u_y=2xe^y $$

I solved this equation like following:

Divided the equation by $x$: $$ u_{xy}+ \frac {1}{x}u_y = 2e^y $$

Then integrated with respect to $x$: $$ u_y + \ln(x)u_y = 2xe^y+f(y) $$

Then: $$ u_y (1+\ln(x)) = 2xe^y +f(y)$$

Then integrated with respect to $y$: $$ u = \frac{2x}{1+\ln(x)} e^y + \int f(y) + h(y)$$

However I am thinking that my 3rd step is incorrect. Can you help guys, if I am doing something wrong?

Robert Howard
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Alpys Rauan
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1 Answers1

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In fact the second step is incorrect: Since $u$ may depend on $x$, integrating $f_x u_y$ in general does not give $f u_y$.

Hint We can proceed as follows: Since $u$ only appears in the equation differentiated by $y$, we can produce a lower-order equation by writing $v := u_y$: $$x v_x + v = 2 x e^y .$$

Additional hint The left-hand side is just $(x v)_x$, so we can immediately integrate to obtain $x v$, thus $v$, and then integrate again to obtain $u$.

Travis Willse
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  • thanks for your response, I integrated (xv)_x by x and obtained something like that: xv=(x^2)(e^y)+g(y), then i solely get v=x(e^y)+(1/x)g(y), then since v=u_y, I am just integrating by y and I got: u=xe^y + (1/x)integral(h(y)) + g(x) . I hope it is correct. – Alpys Rauan Mar 12 '19 at 05:55
  • You're welcome, I hope you found it useful. Remember, you can always check a solution by substituting back in the original equation. – Travis Willse Mar 12 '19 at 06:05